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I am trying to show $D_{6}$ is isomorphic to $S_3$ by showing a function defined by

$\phi(r)= (1 2 3)$ and $\phi(s) = (2 3)$ is homomorphism and bijective.

I see that the $ker(\phi)=\{1\}$. And hence since the two groups are finite it is bijective.

I am having trouble showing that it is homomorphism.

I must show that $\phi(g_1g_2) = \phi(g_1)\phi(g_2)$.

In general, do I need to check all possible combination of $g_1g_2$ or is there a easier way to see this?

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$D_6$ is a subset of $S_3$ with six elements. So actually they are equal.

If you are using some other definition of $D_6$ that doesn't display them as permutations, then I'm sure you'll have no trouble labeling the vertices of a triangle with 1,2,3 and identifying the symmetries with elements of $S_3$.

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You are already done. A group homomorphism is determined by the values on generators. Since $r,s$ generate $D_6$, the definition $\phi(r)=(123)$ and $\phi(s)=(23)$ gives all other values, e.g., $\phi(rs)=(123)(23)=\phi(r)\phi(s)$, $\phi(s^2)=(23)(23)=id$, etc.

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Let $G$ be a nonabelian group of order $6$. By Cauchy's theorem there is an element $x\in G$ of order $2$ and an element $y\in G$ of order $3$. Since $[G:\langle y\rangle ]=2$, the subgroup $\langle y\rangle $ is normal. In particular the conjugate $xyx^{-1}$ must lie in $y$. It cannot be $1$ since $y\neq 1$, and it cannot be $y$ since $x$ and $y$ do not commute for our group is nonabelian. So it must be $y^{-1}$, for $\langle y\rangle =\{1,y,y^{-1}\}$, that is (since $x^2=1$), $xyxy=1$. Hence $G= \langle x\rangle \ltimes \langle y\rangle$, and it follows that $G$ is presented as the group $$G=\langle x,y\mid x^2,y^3,(xy)^2\rangle $$

so there is only one nonabelian group of order $6$, up to isomorphism. In particular $S_3$ and $D_6$, which are both nonabelian groups of order $6$, must be isomorphic to $G$, and hence isomorphic themselves.

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